#int_1^xf(t)dt=x(f(x))^2# deriving both sides

#f(x)=(f(x))^2+2xf(x)f'(x)# or

#(2xf'(x)+f(x)-1)f(x)=0# Supposing that #f(x) ne 0# we have

#2xf'(x)+f(x)-1=0# Solving this linear differential equation we obtain

#f(x)=1+C_1/sqrt(x)# The constant is determined submiting this solution into the first relationship giving

#(x + 2(sqrt[x]-1)C_1-1)/x=(1+C_1/sqrtx)^2# giving #C_1=-1# so the function is

#f(x)=1-1/sqrtx# and also #f(x)=0#